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\begin{document}
\begin{itemize}
\item

Recall that in the transfer matrix approach, we start with a partition function

$$Z_N = \sum_{\sigma_1} \sum_{\sigma_2} \cdots \sum_{\sigma_N} e^{w(\sigma_1,\sigma_2)} e^{w(\sigma_2,\sigma_3)} \cdots e^{w(\sigma_N,\sigma_1)}$$

and contrive a matrix $T$ such that

$$e^{w(\sigma,\sigma')} = \ang{ \sigma | T | \sigma' }$$

then

$$Z_N = \sum_{\sigma_1} \sum_{\sigma_2} \cdots \sum_{\sigma_N} \ang{ \sigma_1 | T | \sigma_2 } \ang{ \sigma_2 | T | \sigma_3 } \cdots \ang{ \sigma_N | T | \sigma_1 } = \sum_{\sigma_1} \ang{ \sigma_1 | T^N | \sigma_1 } = \mathrm{tr} \, T^N$$

Now in a decimation procedure with e.g. $\lambda = 2$ we have

$$Z_N = \sum_{\sigma_1} \sum_{\sigma_3} \sum_{\sigma_5} \cdots \sum_{\sigma_N} \left[ \sum_{\sigma_2} \ang{ \sigma_1 | T | \sigma_2 } \ang{ \sigma_2 | T | \sigma_3 } \right] \left[ \sum_{\sigma_4} \ang{ \sigma_3 | T | \sigma_4} \ang{ \sigma_4| T | \sigma_5 } \right] \cdots $$
$$= \sum_{\sigma_1} \sum_{\sigma_3} \sum_{\sigma_5} \cdots \sum_{\sigma_N} \ang{ \sigma_1 | T^2 | \sigma_3 } \ang{ \sigma_3 | T^2 | \sigma_5 } \cdots$$ 
$$= \sum_{\sigma_1} \sum_{\sigma_3} \sum_{\sigma_5} \cdots \sum_{\sigma_N} \ang{ \sigma_1 | T' | \sigma_3 } \ang{ \sigma_3 | T' | \sigma_5 } \cdots$$ 

so the RG equation in this case is $\ang{ \sigma | T' | \sigma' } = \ang{ \sigma | T^2 | \sigma' }$, or indeed for more general integer $\lambda$

$$\ang{ \sigma | T' | \sigma' } = \ang{ \sigma | T^\lambda | \sigma' }$$

\item

For the 1D Ising in an external field

$$T = \left(\begin{array}{cc} 1/(uv) & u \\u & v/u \end{array}\right) \implies T^2 = \left(\begin{array}{cc} 1/(uv)^2 + u^2 & 1/v + v \\ 1/v + v & u^2 + v^2/u^2 \end{array}\right)$$

where we have defined $u := \exp(-K)$, $v := \exp(-h)$. The RG equations are thus

$$u' = \frac{1}{v} + v, \, \frac{1}{u'v'} = \frac{1}{(uv)^2} + u^2, \, \frac{v'}{u'} = \frac{v^2}{u^2} + u^2$$

we wish to relate $u'$ with $u$ and $v'$ with $v$, which can be achieved by coupling these equations appropriately

$$u' \frac{v'}{u'} = \left(\frac{1}{v} + v\right) \left( \frac{v^2}{u^2} + u^2 \right) \implies v' = \frac{v}{u^2} + \frac{u^2}{v} + \frac{v^3}{u^2} + v^2$$

$$u' \frac{1}{u'v'} = \left(\frac{1}{v} + v\right) \left( \frac{1}{(uv)^2} + u^2 \right) \implies v' = \frac{v}{u^2} + \frac{u^2}{v} + \frac{v^3}{u^2} + v^2$$


\end{itemize}
\end{document}